Nitrogen has oxides like $\ce{N2O}$, $\ce{NO}$, $\ce{NO2}$, $\ce{N2O3}$, $\ce{N2O5}$, but as far as I know, phosphorus has probably only $\ce{P2O3}$ and $\ce{P2O5}$. Why does nitrogen have so many oxides in the first place? Also why doesn't phosphorus have the analogous oxides?
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1$\begingroup$ See en.wikipedia.org/wiki/Diphosphorus_tetroxide, also pl.wikipedia.org/wiki/Tlenki_fosforu $\endgroup$– MithoronCommented Mar 6, 2017 at 17:19
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2 Answers
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Also the oxides for phosphorus are somewhat special and I think they just have a high tendendy to form. Organic chemistry like the Wittig-reaction uses the oxophilia of phosphorus a lot.
If you assume from basic organic chemistry that 6-membered rings are something stable and a cyclopropane isn't that stable you can try to apply that to phosphorus as well. But it is a different element and you'll come up with a basic thumb rule that says that 5-membered rings are quite good for phosphorus while the more 3-membered rings you have the more unstable it will become. $\ce{P4}$, which is the smallest (stable) unit of phosphorus however consists of triangles, only so it has to be highly reactive. the bonding orbitals just can't bent that way to get the tetrahedral angles and the bonds are sometimes called 'bananoid' bonds for this reason. If oygen is present it will 'insert' into this $\ce{P-P}$ bond and act as a bridge for it that way you loose a lot of ring strain on the P atoms. Your tetrahedron has 6 of these edges and all of them are unstable so it will autoamtically form a $\ce{P4O6}$ or $\ce{P2O3}$ as you know it. And if you check on its structure you will see that it is only a tetrahedron where each edge is capped by an oxide. For the next step you will have to descide whether P is nucleophilic or electrophilic and this depends on the other partner in the reaction. With oxygen however it is further oxidized to form $\ce{P4O10}$ or $\ce{P2O5}$.
And phosphorus is just so reactive towards oxygen that this happens for all the possible positions on the tetrahedron therefore those two oxides are the most favored ones and will likely always form.
Nitrogen on the other hand has different structural motifs starting of with an $\ce{N2}$ molecule for example and that $\ce{N \tbond N}$ bond is really strong, too. So the way how it is oxidized is different from how phosphorus behaves.
I don't know if similar oxides exist for phosphorus as we know them for nitrogen and this was only a guess based on a phosphorus lecture I heard a while ago.
answered Mar 6, 2017 at 17:38
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It is generally due to the larger size of phosphorus. Nitrogen and oxygen both are members of 2nd period.Hence the p$\pi$ - p$\pi$ bond formed between nitrogen and oxygen has larger area of overlap resulting in shorter bond length and increased stability. Phosphorus is a member of 3rd period so when 3p orbital of phosphorus overlaps with 2p orbital of oxygen, due to difference in energy and size the area of overlap decreases which leads to lesser stability.Hence it forms less oxides.
